Given two numbers N, M in the bases X, Y and another base P. The task is to find the product of N and M and represent the product in base P.
Examples:
Input: N = 101, M = 110, X = 2, Y = 2, P = 16
Output:1E
Explanation: NX * MY = (101)2 * (110)2 = (11110)2
(11110)2 = (1E)16Input: N = 101, M = A, X = 2, Y = 20, P = 16
Output: 32
Explanation: Nx = (101)2 = (5)20
NX * MY = (5)20 * (A)20 = (2A)20
(2A)20 = (32)16
Approach: The approach is to convert the given numbers in decimal, perform the product and then turn it back to a number of base p. Follow the steps mentioned below:
- Convert NX and MY to decimal number.
- Perform multiplication on the decimal numbers.
- Convert the result of multiplication from decimal to base P.
Below is the implementation of the above approach.
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Convert Number from a given base// to decimal// Return the value of a char.static int value(char c){ if (c >= '0' && c <= '9') return (int)c - '0'; else return (int)c - 'A' + 10;}// Function to convert a// number from given base to decimalint toDecimal(string s, int base){ int length = s.length(); // Initialize power of base and result int power = 1, ans = 0; // Decimal equivalent of s for (int i = length - 1; i >= 0; i--) { ans += value(s[i]) * power; power = power * base; } return ans;}// Function to convert decimal// to any given basechar reverseValue(int n){ if (n >= 0 && n <= 9) return (char)(n + 48); else return (char)(n - 10 + 65);}// Function to convert a given// decimal number to a base 'base'string toBase(int base, int num){ string s = ""; // Convert input number is given // base by repeatedly dividing it // by base and taking remainder while (num > 0) { s += reverseValue(num % base); num /= base; } string sb = ""; // Append a string into StringBuilder sb += (s); // Reverse the result reverse(sb.begin(), sb.end()); return sb;}// Function to find// the product of N and Mvoid findProduct(string N, int X, string M, int Y, int P){ // Convert N to decimal int decimalX = toDecimal(N, X); // Convert y to decimal int decimalY = toDecimal(M, Y); // Multiply the decimal numbers int product = decimalX * decimalY; // Convert product to base string result = toBase(P, product); // Print the result cout << (result);}// Driver codeint main(){ string N = "101", M = "110"; int X = 2, Y = 2, P = 16; findProduct(N, X, M, Y, P); return 0;}// This code is contributed by Potta Lokesh |
Java
// Java code to implement above approachimport java.io.*;class GFG { // Function to find // the product of N and M static void findProduct(String N, int X, String M, int Y, int P) { // Convert N to decimal int decimalX = toDecimal(N, X); // Convert y to decimal int decimalY = toDecimal(M, Y); // Multiply the decimal numbers int product = decimalX * decimalY; // Convert product to base String result = toBase(P, product); // Print the result System.out.println(result); } // Convert Number from a given base // to decimal // Return the value of a char. static int value(char c) { if (c >= '0' && c <= '9') return (int)c - '0'; else return (int)c - 'A' + 10; } // Function to convert a // number from given base to decimal static int toDecimal(String s, int base) { int length = s.length(); // Initialize power of base and result int power = 1, ans = 0; // Decimal equivalent of s for (int i = length - 1; i >= 0; i--) { ans += value(s.charAt(i)) * power; power = power * base; } return ans; } // Function to convert decimal // to any given base static char reverseValue(int n) { if (n >= 0 && n <= 9) return (char)(n + 48); else return (char)(n - 10 + 65); } // Function to convert a given // decimal number to a base 'base' static String toBase(int base, int num) { String s = ""; // Convert input number is given // base by repeatedly dividing it // by base and taking remainder while (num > 0) { s += reverseValue(num % base); num /= base; } StringBuilder sb = new StringBuilder(); // Append a string into StringBuilder sb.append(s); // Reverse the result return new String(sb.reverse()); } // Driver code public static void main(String[] args) { String N = "101", M = "110"; int X = 2, Y = 2, P = 16; findProduct(N, X, M, Y, P); }} |
Python3
# Python code for the above approach# Convert Number from a given base# to decimal# Return the value of a char.def value(c): if (ord(c[0]) >= ord('0') and ord(c) <= ord('9')): return int(c) else: return ord(c[0]) - ord('A') + 10# Function to convert a# number from given base to decimaldef toDecimal(s,base): length = len(s) # Initialize power of base and result power,ans = 1,0 # Decimal equivalent of s for i in range(length-1,-1,-1): ans += value(s[i]) * power power = power * base return ans# Function to convert decimal# to any given basedef reverseValue(n): if (n >= 0 and n <= 9): return chr(n + 48) else: return chr(n - 10 + 65)# Function to convert a given# decimal number to a base 'base'def toBase(base, num): s = "" # Convert input number is given # base by repeatedly dividing it # by base and taking remainder while (num > 0): s += reverseValue(num % base) num = (num//base) sb = "" # Append a string into StringBuilder sb += (s) # Reverse the result sb = sb[::-1] return sb# Function to find# the product of N and Mdef findProduct(N, X, M, Y, P): # Convert N to decimal decimalX = toDecimal(N, X) # Convert y to decimal decimalY = toDecimal(M, Y) # Multiply the decimal numbers product = decimalX * decimalY # Convert product to base result = toBase(P, product) # Print the result print(result)# Driver codeN, M = "101", "110"X, Y, P = 2, 2, 16findProduct(N, X, M, Y, P)# This code is contributed by shinjanpatra |
C#
// C# code to implement above approachusing System;class GFG{ // Function to find // the product of N and M static void findProduct(String N, int X, String M, int Y, int P) { // Convert N to decimal int decimalX = toDecimal(N, X); // Convert y to decimal int decimalY = toDecimal(M, Y); // Multiply the decimal numbers int product = decimalX * decimalY; // Convert product to base String result = toBase(P, product); // Print the result Console.Write(result); } // Convert Number from a given base // to decimal // Return the value of a char. static int value(char c) { if (c >= '0' && c <= '9') return (int)c - '0'; else return (int)c - 'A' + 10; } // Function to convert a // number from given base to decimal static int toDecimal(String s, int _base) { int length = s.Length; // Initialize power of base and result int power = 1, ans = 0; // Decimal equivalent of s for (int i = length - 1; i >= 0; i--) { ans += value(s[i]) * power; power = power * _base; } return ans; } // Function to convert decimal // to any given base static char reverseValue(int n) { if (n >= 0 && n <= 9) return (char)(n + 48); else return (char)(n - 10 + 65); } // Function to convert a given // decimal number to a base 'base' static String toBase(int _base, int num) { String s = ""; // Convert input number is given // base by repeatedly dividing it // by base and taking remainder while (num > 0) { s += reverseValue(num % _base); num /= _base; } String sb = ""; // Append a string into StringBuilder sb += s; // Reverse the result char[] arr = sb.ToCharArray(); Array.Reverse(arr); return new string(arr); } // Driver code public static void Main() { String N = "101", M = "110"; int X = 2, Y = 2, P = 16; findProduct(N, X, M, Y, P); }}// This code is contributed by saurabh_jaiswal. |
Javascript
<script>// JavaScript code for the above approach// Convert Number from a given base// to decimal// Return the value of a char.function value(c){if (c.charCodeAt(0) >= '0'.charCodeAt(0) && c.charCodeAt(0) <= '9'.charCodeAt(0)) return parseInt(c);else return c.charCodeAt(0) - 'A'.charCodeAt(0) + 10;}// Function to convert a// number from given base to decimalfunction toDecimal(s,base){let length = s.length;// Initialize power of base and resultlet power = 1, ans = 0;// Decimal equivalent of sfor (let i = length - 1; i >= 0; i--){ ans += value(s[i]) * power; power = power * base;}return ans;}// Function to convert decimal// to any given basefunction reverseValue(n){if (n >= 0 && n <= 9) return String.fromCharCode(n + 48);else return String.fromCharCode(n - 10 + 65);}// Function to convert a given// decimal number to a base 'base'function toBase(base, num){let s = "";// Convert input number is given// base by repeatedly dividing it// by base and taking remainderwhile (num > 0){ s += reverseValue(num % base); num = Math.floor(num/base);}let sb = "";// Append a string into StringBuildersb += (s);// Reverse the resultsb = sb.split("").reverse().join("");return sb;}// Function to find// the product of N and Mfunction findProduct(N, X, M, Y, P){// Convert N to decimallet decimalX = toDecimal(N, X);// Convert y to decimallet decimalY = toDecimal(M, Y);// Multiply the decimal numberslet product = decimalX * decimalY;// Convert product to baselet result = toBase(P, product);// Print the resultdocument.write(result,"</br>");}// Driver codelet N = "101", M = "110";let X = 2, Y = 2, P = 16;findProduct(N, X, M, Y, P);// This code is contributed by shinjanpatra</script> |
Output
1E
Time Complexity: O(D) where D is the maximum number of digits in N, M and product
Auxiliary Space: O(1)
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